September 23, 2025
Let \({\mathcal{P}}\) be the set of prime numbers, and \(X_p,\, p\in{\mathcal{P}}\) be a sequence of independent random variables such that \({\mathbb{P}}(X_p=\pm 1)=1/2\). Let \(({\theta}_j)_{j=1}^{\infty}\) the corresponding random multiplicative functions of Radamacher type, namely, \({\theta}_j=\prod_{p|j}X_p\) if \(j\) is square free and \({\theta}_j=0\) otherwise. The motivation behind considering these variables comes from the Riemann Hypothesis since \(({\theta}_\ell)\) can be viewed as the random counterpart of the Möbius function \(\mu(\ell)\) and the RH is equivalent to \(\sum_{\ell\leq n}\mu(\ell)=o(n^{1/2+\varepsilon}), \forall \varepsilon>0\). Denote \(S_n=\sum_{\ell=1}^n{\theta}_\ell\). It is a natural guiding conjecture that \(S_n/\sqrt n\) obeys the central limit theorem (CLT). However, S. Chatterjee conjectured (as expressed in ) that the CLT should not hold. Chatterjee’s conjecture was proved by Harper , and by now it is a direct consequence of a more recent breakthrough by Harper [@Har20] that \(\frac{S_n}{\sqrt n}\to 0\) in distribution. Nevertheless, the question whether there exists a sequence \(a_n=o(\sqrt n)\) such that \(S_n/a_n\) converges to some limit remains a mystery. In this paper we, in particular, show that the answer is negative. Our proof is based on trigonometric identities, Levi’s continuity theorem, the prime numbers theorem and the growth rates of the prime omega function \({\omega}(n)\). Combining these tools we show that if a weak limit exists then the characteristic function of \(S_n/a_n\) must converge to \(0\) at any nonzero point. As a byproduct of our method we are also able to provide essentially optimal upper bounds on high order moments similar to [@Har19], using different methods and for more general \(X_p\)’s.
The study of mean values of multiplicative functions has played a central role in analytic number theory for hundreds of years. At the heart of the subject lies the Möbius function \(\mu(n)\) which is the multiplicative function supported on square free (natural) numbers and defined to be \(-1\) on the primes. The associated generating series is \(\sum_n\mu(n)/n^s\) which is absolutely convergent for \(\text{Real}(s)<1\), and is equal to \(1/\zeta(s)\), where \(\zeta\) denotes the Riemann zeta-function. It is this connection with the Riemann zeta-function which makes the study of partial sums of \(\mu(\cdot)\) so intriguing; indeed, the Riemann Hypothesis is equivalent to the estimate \[\label{40146141} \sum_{\ell\leq n}\mu(n)\ll n^{\frac{1}{2}+\varepsilon}\tag{1}\] for all \(\varepsilon\). This kind of growth is similar to the almost sure growth rates of partial sums \(\sum_{\ell=1}^n {\theta}_\ell\) of independent and identically distributed (iid) zero mean random variables \(({\theta}_\ell)\). In fact the law of iterated logarithm guarantees the growth rate \(\sqrt{n\ln(\ln(n))}\). The above estimate is often framed in terms of the “pseudo-randomness” of the Möbius function (e.g., [@IK04]). In this paper we study this pseudo-random behavior through the study of so-called random multiplicative functions, as discussed in the next paragraphs.
A naive heuristic which can be used to understand why 1 should be true is to model the Möbius function by a sequence of iid random variables \({\theta}_\ell, \ell\in\text{SF}\) taking the values \(\pm 1\) with equal probability, where \(\text{SF}\) is the set of square free positive integers. If this were the case, the partial sums \(\sum_{\ell=1}^n\mu(\ell)\) would mimic a random walk with mean zero and variance equal to number of squarefree integers up to \(n\) (which is of order \(n/\zeta(2)=6n/\pi^2\)); in particular, we would expect that 1 should hold.
In [@Le31], Lévy objected to the above model, as the sequence of random variables \(({\theta}_\ell), \ell\in\text{SF}\) lacks the multiplicative structure inherent in \((\mu(n))\). In an attempt to rectify this, in 1944 Wintner [@Win44] introduced the concept of random multiplicative functions. More precisely Wintner introduced Rademacher random multiplicative functions to model the behavior of the Möbius function \(\mu(n)\). These random variables are defined by \({\theta}_\ell=\prod_{p|\ell}X_p\) if \(\ell\in\text{SF}\) and \({\theta}_\ell=0\) otherwise. Here \((X_p)\) is an iid sequence such that \({\mathbb{P}}(X_p=\pm1)=1/2\). Note that \(({\theta}_\ell)\) is a random model for \((\mu(\ell))\), as \((\mu(\ell))\) is simply a specific realization of \(({\theta}_\ell)\).
Let us provide some details concerning Wintner’s results. Using the theory of Dirichlet series, Wintner showed that \(\sum_{\ell}{\theta}_\ell\ell^{-s}\) is is almost always convergent for \(\text{Real}(s)>1/2\). He further showed that, for all \(\varepsilon>0\), both \(\sum_{\ell\leq n}{\theta}_\ell=O(n^{1/2+\varepsilon})\) and \(\sum_{\ell\leq n}{\theta}_\ell\not=O(n^{1/2-\varepsilon})\) hold almost everywhere, leading to the egregious statement that “the Riemann Hypothesis is almost surely true”. The study of random multiplicative functions has flourished in recent years, most notably with work of Harper (e.g., [@Har13; @HNR15; @Har19; @Har20; @Har21; @Har23]). We refer the reader to [@Har13] and the introduction of [@Har19], [@Har21] for a meticulous overview of this subject.
A classical question of interest which attracted a lot of attention is to understand the distribution and the sizes of the partial sums \(S_n=\sum_{\ell=1}^n{\theta}_\ell\). It is a natural guiding conjecture that the central limit theorem (CLT) holds, namely that \[S_n/\sqrt n\to{\mathcal{N}}(0,a),\,\text{ as }\,n\to\infty\] in distribution, where \(a\) is the density of \(\text{SF}\) (\(a=\frac{6}{\pi^2}\)) and \({\mathcal{N}}(0,a)\) is the normal distribution with zero mean and variance \(a\). However S. Chatterjee suggested that this should not hold. Chatterjee’s conjecture (expressed in [25]), was proved by Harper , using an intricate conditioning argument. It is now a direct consequence of a more recent breakthrough work by Harper [@Har20] on Helson’s conjecture that in fact \[\lim_{n\to\infty}\frac{S_n}{\sqrt n}=0\,\,\text{ in distribution.}\] Since the limit \(0\) is constant we automatically get convergence in probability to \(0\). This result in a sense is very counter intuitive since \(\|n^{-1/2}S_n\|_{L^2}^2\asymp \frac{6}{\pi^2}\) and so the convergence to \(0\) does not hold in \(L^2\). Moreover, for sums of independent (or weakly dependent) random variables the linear growth of the variance implies the CLT. Interestingly, if one restricts to several natural subsums, Chatterjee and Soundararajan , Harper and Hough and Soundararajan and Wu [@CLT] established central limit theorems. In [@Gor] Gorodetsky and Wong established the CLT for weighted sums \(\sum_{\ell=1}^nf(\ell){\theta}_\ell\) when \(f(\ell)\) is a multiplicative function exhibiting some (averaged) decay conditions as \(\ell\to\infty\), which excludes the classical case when \(f(\ell)=1\). It is a deep mystery whether appropriately normalized partial sums \(S_n\) have a limiting distribution as \(n\to\infty\).
The fundamental difficulty in proving the CLT stems from the fact that the values \({\theta}_n\) and \({\theta}_m\) are not independent whenever \(\text{gcd}(m,n)>1\). It is also not hard to show that the sequence \(({\theta}_\ell)\) is not weakly dependent, and instead it exhibits strong long range dependence. For instance, if \(\ell\) is an odd square free number then \({\theta}_{2\ell}={\theta}_{\ell}X_2\), and more complicated nontrivial long range dependence can be described. The sequence \(({\theta}_\ell)\) also does not have a weakly locally dependent graph structure (see [@HK])), which makes Stein’s method effective. Therefore, it seems unlikely that the corresponding sums \(S_n\) can be directly treated using classical tools from probability that involve some kind of weak dependence.
In this paper we will resolve this problem and show that if \(\liminf_{n\to\infty}a_n>0\) and \(a_n=o(\sqrt n)\) then \(S_n/a_n\) does not converge in distribution (not even along a subsequence). In fact, we will show that the above statement is true for much more general random multiplicative functions of Radamacher type. In a sense this shows that the conditions in [@Gor] are close to optimal since without some decay of the weights there could be no weak convergence. As a byproduct of our method we are also able to provide essentially optimal upper bounds on high order moments of \(S_n\) similar to [@Har19], using different methods and for more general classes of Rademacher functions. The question concerning the completely multiplicative case (i.e. the Steinhaus setting) when \(X_p\) are distributed uniformly on the unit circle and \({\theta}_\ell=\prod_{j=1}^d(X_{p_j})^{e_j},\, \ell=\prod_{j=1}^{d}p_j^{e_j}\) remains open. As this paper demonstrates, the long range dependence described above is a true obstruction to any type of weak convergence (and not only to the CLT) under any magnitude of normalization \(a_n\) beyond the one which ensures a zero limit (i.e. when \(\liminf_{n\to\infty}\frac{a_n}{\sqrt n}>0\)).
To appreciate the peculiarity of our results and to provide a more complete picture we will briefly describe a related class of problems. Let \(P\in{\mathbb{Z}}[x]\) be a nonlinear polynomial and define \(Y_\ell={\theta}_{P(\ell)}\). Then, as opposed to the case of linear polynomials the partial sums \(\sum_{\ell=1}^n Y_\ell\) obey the CLT for a wide range of nonlinear polynomials, see [@GAFA; @TAMS; @Mub] and references therein. Since this paper is focused on linear times and Rademacher functions we prefer not to elaborate too much on the subject, but let us mention that the motivation behind considering polyomial time comes from the Chowla conjecture (see again [@GAFA; @TAMS; @Mub]). This conjecture remains unsolved (see [@Tao] for a proof of an averaged version) but the results in [@GAFA; @TAMS; @Mub] can be viewed as a proof of a random counterpart of the conjecture (aka the “random Chowla conjecture"), and they provide evidence to the validity of the deterministic conjecture. We believe that the ideas in the paper are likely to yield the CLT in the polynomial case to the more general Random Radememacher functions considered in this paper, but we will address this problem in a different venue. The motivation behind considering more general \(X_p\)’s comes from discovering new patterns in the asymptotic behavior of prime numbers. Indeed, the corresponding random multiplicative functions can model more complicated multiplicative deterministic functions, and the CLT in the random case can provide evidence to an appropriate deterministic results. Of course, this is also true for linear polynomials but in that case there is no weak limit unless \(a_n\geq c\sqrt n\).
In this section we will consider more general independent random variables \(X_p\) than the ones discussed in Section 1. All of our conditions will trivially hold true in the classical Rademacher case when \({\mathbb{P}}(X_p=\pm1)=1/2\). However, already in the iid case our conditions include a wide class of bounded symmetric random variables.
Let us begin with a description our our setup. Let \({\mathcal{P}}\) be the set of prime numbers and let \(X_p, p\in{\mathcal{P}}\) be an independent (not necessarily identically distributed) sequence of symmetric non-constant random variables such that \[\label{2323} M:=\sup_{p\in{\mathcal{P}}}\|X_p\|_{L^\infty}<\infty \,\text{ and }\,\inf_{p\in{\mathcal{P}}}{\mathbb{E}}[(X_p)^2]>0.\tag{2}\]
Definition 1. The random multiplicative functions \({\theta}_\ell\) of Rademacher type generated by \((X_p)\) are defined by \({\theta}_\ell=\prod_{p|\ell}X_p\) when \(\ell\) is square free, and otherwise \({\theta}_\ell=0\).
The classical Rademacher setting concerns the case when \({\mathbb{P}}(X_p=\pm1)=1/2\), see the discussion in Section 1, but here we can consider more general random variables \(X_p\). Set \[S_n=\sum_{\ell=1}^n{\theta}_\ell.\]
Notice that \({\theta}_\ell\) are orthogonal since \(X_p\) have zero mean and two distinct square free integers cannot have the same prime factors. Therefore, \[\|S_n\|_{L^2}^2=\sum_{\ell=1}^n{\mathbb{E}}[({\theta}_\ell)^2]=\sum_{\ell\in\text{SF}\cap N_n}\prod_{p|\ell}{\mathbb{E}}[(X_p)^2]\] where \(\text{SF}\) is the set of all square free integers and \(N_n=\{1,2,...,n\}\).
Lemma 1. Let \(a\leq 1\leq b\) and suppose that \(a\leq {\mathbb{E}}[(X_p)^2]\leq b\) for all \(p\). Then there exist absolute constants \(c,C_1,C_2>0\) such that for all \(n\) large enough we have \[C_1n^{1-\frac{c|\ln a|}{\ln(\ln(n))}}\leq \|S_n\|_{L^2}^2\leq C_2n^{1+\frac{c\ln b}{\ln(\ln(n))}}.\]
Proof. Let \({\omega}(\ell)\) is the prime omega function which assigns to \(\ell\) the number of distinct prime factors. Then \({\theta}_\ell\) is a product of \({\omega}(\ell)\) independent random variables \(X_{p_i}\). Taking into account the above formula for \(\|S_n\|_{L^2}^2\) the lemma follows from the following facts. The first one is that (see [@Mir]) the density of the square free numbers is \(6/\pi^2\). The second one is that (see [@omega]) \({\omega}(\ell)\leq \frac{c\ln(\ell)}{\ln(\ln \ell)}\) for some constant \(c\) and all \(\ell\) large enough. The third fact is that the function \(g(x)=\frac{\ln x}{\ln(\ln (x))}\) is increasing on some ray \([u,\infty), u>3\). ◻
We note that the lemma shows that \(\|S_n\|_{L^2}^2\) is almost linear in \(n\) in the sense that it is of order \(n^{1\pm\varepsilon_n}\) for some sequence \(\varepsilon_n\to0\). However, we still have \(\lim n^{\varepsilon_n}=\infty\), but at a very slow rate.
Corollary 1. If \(\inf_{p}{\mathbb{E}}[(X_p)^2]\geq1\) then \(V:=\limsup_{n\to\infty}\frac{1}{n}\|S_n\|_{L^2}^2>0.\) Moreover, if \({\mathbb{E}}[X_p^2]=1\) for all \(p\) then \[V=\lim_{n\to\infty}\frac{1}{n}\|S_n\|_{L^2}^2=\frac{\pi^2}{6}.\]
When \(V>0\) and we will show that \(S_n/a_n\) cannot converge when \(a_n=o(\sqrt n)\). This includes the classical Rademacher case when \({\mathbb{P}}(X_p=\pm1)=1/2\) and then the limit superior is an actual limit and \(V=\frac{6}{\pi^2}\). Another example when \(V>0\) is when \(X_p\) are uniformly distributed on \([-a_p,a_p]\) with \(\sup_pa_p<\infty\) and \(a_p\geq \sqrt 3\). Taking \(a_p=\sqrt 3\) we get that the limit exists and equals \(\frac{6}{\pi^2}\), and many other examples can be given.
Now we are ready to state our first main result.
Theorem 1. Let \(X_p, p\in{\mathcal{P}}\) be a sequence of symmetric independent random variables satisfying 2 . Then, there exists no sequence \((a_n)\) such that \(\liminf_{n\to\infty}a_n>0\), \(a_n=o(\|S_n\|_{L^2})\) and \(S_n/a_n\) converges in distribution (not even along a subsequence).
In particular when \(X_p\) are iid, symmetric non-constant, bounded and \({\mathbb{E}}[|X_2|^2]\geq 1\) then there is no weak limit when \(a_n=o(\sqrt n)\).
As discussed before the condition \(a_n=o(\|S_n\|_{L^2})\) is optimal since when \({\mathbb{P}}(X_p=\pm1)=1/2\) we have that \(S_n/\sqrt n\to 0\) and \(\|S_n\|_{L^2}\asymp \frac{6n}{\pi^2}\). Notice also that the condition \(\liminf_{n\to\infty}a_n>0\) cannot be improved for otherwise we could take \(a_n\to0\) fast enough to ensure that \(S_n/a_n\to\infty\) in an appropriate sense. Our next result concerns the magnitude of high moments of \(S_n\) that were already studied by Harper in the classical Rademacher case (see [@Har19] and references therein).
Theorem 2. Suppose \(M=\sup_{p}\|X_p\|_{L^\infty}<\infty\). Then there exists \(c>0\) such that for every \(r\in{\mathbb{N}}\) and \(n\in{\mathbb{N}}\) we have \[{\mathbb{E}}[(S_n)^{2r}]\leq C_r n^r(\ln(n))^{2r}\left(\max(1,M)\right)^{c\frac{\ln(n)}{\ln(\ln(n))}}\] where \(C_r\) is a constant that depends only on \(r\).
In fact, using trigonometric identities we obtain an explicit close formula 14 for \({\mathbb{E}}[(S_n)^{2k}]\), from which Theorem 1 follows. Note that \(M^{\frac{2kc\ln(n)}{\ln(\ln(n))}}=o(n^\varepsilon)\) for every \(\varepsilon>0\). Using this fact and Theorem 2 the following result follows.
Corollary 2. For every \(\varepsilon>0\) there exists a random variable \(B_{\varepsilon}\in L^{[\frac{1}{2\varepsilon}]+1}\) such that for all \(n\), \[|S_n|\leq B_{\varepsilon}n^{1/2+\varepsilon}, \,\text{ almost surely}.\]
Proof. Let \(\varepsilon>0\). Then for all \(k>[1/(2\varepsilon)]+1\) have \[\left\|\sup_{n\in{\mathbb{N}}}(n^{-1/2-\varepsilon}|S_n|)\right\|_{L^{2k}}^{2k}\leq\sum_{n\geq1}n^{-k-2k\varepsilon}{\mathbb{E}}[|S_n|^{2k}]<\infty.\] ◻
Note that the above corollary could be a source of hueristically get a better understanding of the structure of the prime numbers by considering multiplicative functions with more general values.
Let us briefly describe the proof’s strategy. For the sake of clarity let us focus on the iid case.
Recall that by Levi’s continuity theorem \(S_n/a_n\) converges in distribution to some random variable \(Z\) if and only if for all real \(t\), \[\lim_{n\to\infty}{\mathbb{E}}[e^{it S_n/a_n}]={\mathbb{E}}[e^{itZ}].\] Using Levi’s theorem we first show that if \(S_n/a_n\) converges in distribution then \(\liminf\frac{a_n^2}{l_n}>0\), where \(l_n=\frac{n}{\ln n}\) (see Lemma 2). The idea is that the random variables \(X_p\) for \(p\in{\mathcal{P}}\) and \(n/2<p\leq n\) appear only once in \(S_n\) and so we can factor the partial sum \(\sum_{n/2<p\leq n}X_p\) out of \(\phi_n(t)={\mathbb{E}}[e^{itS_n/a_n}]\) which together with the prime numbers theorem shows that \(\phi_n(t)\) has a factor \(({\mathbb{E}}[e^{it X_2}])^{cn/\ln n}\) for some \(c>0\). Then we expand \({\mathbb{E}}[e^{it X_2}]\) around the origin. The next step in the proof is to write \[{\mathbb{E}}[e^{it S_n/a_n}]={\mathbb{E}}[\cos(tS_n/a_n)]+i{\mathbb{E}}[\sin(tS_n/a_n)].\] In Proposition 3 we show that for all sufficiently small nonzero \(t\in{\mathbb{R}}\), \[\lim_{n\to\infty}{\mathbb{E}}[\cos(tS_n/a_n)]=0.\] This is done by using a trigonometric identity 4 for expressions of the form \(A=\cos(\sum_{\ell=1}^{n} a_j)\), applied with \(a_\ell=t{\theta}_\ell\), and then using a truncation argument for the resulting sums, showing that it is sufficient to consider a relatively small number of summands in that expansion of \(\cos(tS_n/a_n)\). The low order summands are then bounded using the prime numbers theorem together with a combinatorial argument. This step heavily uses the Rademacher type multiplicativity. The last step is to show that under the assumption that \(S_n/a_n\) converges in distribution (along a subsequence) to some random variable \(Z\) then \[\lim_{n\to\infty}{\mathbb{E}}[\sin(tS_n/a_n)]=0\] which implies that for all nonzero \(t\) we have \({\mathbb{E}}[e^{itZ}]=0\), leading to a contradiction.
Lemma 2. Suppose \(\inf_{p}\|X_p\|_{L^2>0}\) and \(M=\sup_p\|X_p\|_{L^\infty}<\infty\). If there exists a sequence \((a_n)\) such that \(\liminf_{n\to\infty}a_n>0\) and \(S_n/a_n\) converges in distribution then \(\liminf_{n\to\infty}\frac{a_n^2}{l_n}>0\) where \(l_n=\frac{n}{\ln n}\).
Proof. Note that if \(p\) is a prime number such that \(n/2<p\leq n\) then \(X_p\) appears only once in the sum \(S_n\). Indeed, \(X_p\) can only appear in \({\theta}_\ell, \ell\leq n\) if \(p|\ell\), which implies that \(\ell\geq p\). If \(\ell>p\) then \(\ell\) must have another prime factor which means that \(\ell\geq 2p>n\). Let \({\mathcal{P}}_n\) be the set of all prime numbers inside \((n/2,n]\). Set \(s_n=|{\mathcal{P}}_n|\). Then by the prime numbers theorem there are constants \(c_1,c_2>0\) such that for all \(n\) large enough \[c_1\frac{\ln n}{n}\leq s_n\leq c_2\frac{\ln n}{n}\]
Next, denote \(\varphi_p(t)={\mathbb{E}}[e^{itX_p}]\). Now, since \(X_p\) are independent and we see that \[{\mathbb{E}}[e^{itS_n}]=\left(\prod_{p\in{\mathcal{P}}_n}\varphi_p(t)\right){\mathbb{E}}[e^{it(S_n-\sum_{p\in{\mathcal{P}}_n}X_p)}].\] Now, by the Lagrange form of the Taylor remainder of order \(3\) of the function \(\varphi_p(\cdot)\) around the origin we see that for all \(t\) and \(n\), with \({\sigma}_p^2={\mathbb{E}}[(X_p)^2]>0\), \[\label{cos} \left|\varphi_p(t/a_n)-\left(1-\frac{{\sigma}_p^2t^2}{2a_n^2}\right)\right|\leq\frac{|t|^3\cdot {\mathbb{E}}[|X_p|^3]}{6a_n^3}\leq \frac{M|t^3|{\sigma}_p^2}{6a_n^3}\tag{3}\] where the last inequality uses that \({\mathbb{E}}[|X_p|^3]\leq M{\mathbb{E}}[|X_p|^2]\) (by 2 ). Therefore, if \(|t|\) is small enough to ensure that \(\frac{M|t^3|}{6a_n^3}\leq \frac{t^2}{4a_n^2}\), then \[|\varphi_p(t/a_n)|\leq 1-\frac{t^2{\sigma}_p^2}{4a_n^2}\leq e^{-\frac{t^2{\sigma}_p^2}{4a_n^2}}\] where the last inequality uses that \(1-x\leq e^{-x}\) for all \(x\geq 0\). It follows that for all \(n\) large enough and every real nonzero \(t\), \[|{\mathbb{E}}[e^{itS_n/a_n}]|\leq e^{-\frac{t^2}{4a_n^2}\sum_{p\in{\mathcal{P}}_n}{\sigma}_p^2}=e^{-\frac{t^2s_n}{4a_n^2}},\, s_n=\sum_{p\in{\mathcal{P}}_n}{\sigma}_p^2\asymp \frac{n}{\ln n}.\] Then \[|{\mathbb{E}}[e^{itS_n/a_n}]|\leq e^{-\frac{t^2s_n}{4a_n^2}}.\]
Let \(Z\) be the weak limit of \(S_n/a_n\). Then by Levi’s continuity theorem for every \(t\in{\mathbb{R}}\), \[\lim_{n\to\infty}|{\mathbb{E}}[e^{itS_n/a_n}]|={\mathbb{E}}[e^{itZ}].\] Using that \(t\to{\mathbb{E}}[e^{itZ}]\) is continuous, for all \(t>0\) small enough we have \(|{\mathbb{E}}[e^{itZ}]-1|<1/2\) and so for all \(n\) large enough, \[\frac{1}{2}<|{\mathbb{E}}[e^{itS_n/a_n}]|\leq e^{-t^2\frac{s_n}{4a_n^2}}.\] Let us take some sufficiently small nonzero point \(t_0\). Then for all \(n\) large enough we see that \[\frac{t_0^2 s_n}{4a_n^2}\leq \ln 2,\] namely \(a_n^2\geq \frac{t_0^2s_n}{4\ln 2}\). Therefore \(\liminf_{n\to\infty}\frac{a_n}{\sqrt{s_n}}>0\). ◻
Proposition 3. Suppose \(\inf_{p}\|X_p\|_{L^2}>0\) and \(M=\sup_p\|X_p\|_{L^\infty}<\infty\). Let \(l_n=\frac{n}{\ln n}\). Then there exists a constant \({\delta}_0>0\) such that if \(\liminf_{n\to\infty}\frac{a_n^2}{l_n}>0\) and \(\lim_{n\to\infty}\frac{a_n}{\sqrt n}=0\) then for all \(t\in(-\delta_0,\delta_0)\setminus\{0\}\) we have \[\label{UP} \lim_{n\to\infty}{\mathbb{E}}[\cos(tS_n/a_n)]=0.\qquad{(1)}\]
Proof. Recall that for every absolutely convergent series \(A=\sum_{k=1}^\infty a_k\) we have \[\label{Coss} \cos(A)=\sum_{k=0}^{\infty}(-1)^k\sum_{A\subset {\mathbb{N}},\, |A|=2k}\,\,\left(\prod_{\ell\in A}\sin(a_\ell)\prod_{\ell\not\in A}\cos(a_\ell)\right).\tag{4}\] Thus, when applying the above identity with \(a_\ell=t{\theta}_\ell\) for \(\ell\leq n\) and \(a_\ell=0\) otherwise, and using that \(\sin(0)=0\) and \(\cos(0)=1\) we see that with \(N_n=\{1,2,...,n\}\) we have \[\label{Coss1} {\mathbb{E}}[\cos(tS_n)]={\mathbb{E}}\left[\prod_{\ell=1}^n\cos(t{\theta}_\ell)\right]+\sum_{k=1}^{[n/2]}(-1)^k\sum_{A\subset N_n: |A|=2k}{\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell)\right]\tag{5}\] Next, by expanding \(\cos(x)\) around the origin and the Lagrange form of the error term in the Taylor expansion of order \(2\) we see that for all \(t,\ell\) and \(n\), \[\left|\cos(t{\theta}_\ell/a_n)-\left(1-\frac{t^2{\theta}_\ell^2}{2a_n^2}\right)\right|\leq \frac{|t|^3\cdot|{\theta}_\ell|^3}{6a_n^3}.\] Using also that \(\liminf_{n\to\infty}\frac{a_n}{\sqrt{l_n}}>0\) and that \(|{\theta}_\ell|\leq M^{{\omega}(\ell)}\leq M^{c\frac{\ln(n)}{\ln(\ln(n))}}\) we see that for all \(n\) large enough and \(|t|\) small enough we have \[\frac{|t|^3\,\cdot\,|{\theta}_\ell|^3}{6a_n^3}\leq \frac{1}{4} \frac{t^2|{\theta}_\ell|^2}{a_n^2}.\] Therefore, for \(t\) small enough and \(n\) large enough for every \(\ell\leq n\) we have \[\label{Cosss} |\cos(t{\theta}_\ell/a_n)|\leq 1-\frac{({\theta}_\ell)^2t^2}{4a_n^2}\leq e^{-\frac{({\theta}_\ell)^2t^2}{4a_n^2}}\tag{6}\] where the last inequality uses that \(1-x\leq e^{-x}\) for all \(x\geq0\). Hence, \[\left|{\mathbb{E}}\left[\prod_{\ell=1}^n\cos(t{\theta}_\ell)\right]\right|\leq {\mathbb{E}}\left[\exp\left(-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2\right)\right].\] Taking into account our assumption that \(a_n^2=o(\|S_n\|_{L^2}^2)\) we conclude that, \[Z_n:=a_n^{-2}\sum_{\ell=1}^n{\theta}_\ell^2\to \infty\text{ in }L^1\] In particular for \(t\not=0\) since \(a_n^{-2}\sum_{\ell=1}^n{\theta}_\ell^2\geq 0\), possibly along a subsequence, \[-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2\to -\infty\text{ a.s.}\] Indeed, by taking a subsequence \((n_k)\) such that \({\mathbb{E}}\left[Z_{n_k}\right]\geq k^4\) and using the Markov inequality we see that \[{\mathbb{P}}(Z_{n_k}\geq\sqrt{{\mathbb{E}}[Z_{n_k}]})\leq k^{-2}\] and so by the Borel Cantelli lemma \(Z_{n_k}\to\infty\) almost surely. Consequently, by the dominated convergence theorem for every nonzero \(t\), \[\limsup_{n\to\infty}\left|{\mathbb{E}}[\cos(tS_n/a_n)]\right|\leq \lim_{n\to\infty}{\mathbb{E}}\left[e^{-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2}\right]=0.\]
What remains in order to complete the proof of the proposition is to show that for all \(t\in(-1,1)\setminus\{0\}\), \[\label{Showw} \lim_{n\to\infty}\sum_{k=1}^{[n/2]}(-1)^k\sum_{A\subset N_n: |A|=2k}{\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell)\right]=0.\tag{7}\] Next, set \(c_n=\left[n^{1/2+\varepsilon}\sqrt{\ln n}\right]\) for some \(0<\varepsilon<1/4\). We claim that \[\label{Claim} \lim_{n\to\infty}\sum_{k=c_n}^{[n/2]}(-1)^k\sum_{A\subset N_n: |A|=2k}{\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell)\right]=0.\tag{8}\] In order to prove 8 , we note that for all \(n\) large enough, \[\label{sin} |\sin(t{\theta}_\ell/a_n)|\leq |t{\theta}_\ell/a_n|\leq |t/a_n|M^{{\omega}(\ell)}\leq |t/a_n|M^{c\frac{\ln(n)}{\ln(\ln(n))}}:=|t/a_n|M_n.\tag{9}\] Note also that \[\binom{n}{2k}\leq \frac{n^{2k}}{(2k)!}.\] Therefore, \[\sum_{k=c_n}^{[n/2]}\sum_{A\subset N_n: |A|=2k}\left|{\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell/a_n)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell)\right]\right| \leq\sum_{k=c_n}^{[n/2]}\frac{|ntM_n/a_n|^{2k}}{(2k)!}.\] Next, using the Lagrange form of the Taylor remainders of the function \(g(x)=e^x\) we see that for every \(x\geq0\) and \(m\geq0\) we have \[\left|\sum_{k=m+1}^{\infty}\frac{x^k}{k!}\right|=\left|e^x-\sum_{k=0}^{m}\frac{x^k}{k!}\right|\leq\frac{e^x|x|^{m+1}}{(m+1)!}.\] Applying this with \(x=|t|nM_n/a_n\) and \(m=2c_n-1\) for all \(n\) large enough we conclude that \[\sum_{k=c_n}^{[n/2]}\frac{|tnM_n/a_n|^{2k}}{(2k)!}\leq \frac{e^{n|t|M_n/a_n}(nM_n|t|/a_n)^{2c_n}}{(2c_n)!}.\] By combining the above estimate to complete the proof of 8 it is enough to show that \[\lim_{n\to\infty}\frac{e^{n|t|M_n/a_n}(n|t|M_n/a_n)^{2c_n}}{(2c_n)!}=0.\] Let \(\rho>0\). Then \[\frac{e^{n|t|M_n/a_n}(n|t|M_n/a_n)^{2c_n}}{(2c_n)!}<\rho\] is equivalent to \[|t|nM_n/a_n+2c_n\ln (|t|nM_n/a_n)\leq \ln\rho+\ln((2c_n)!).\] Next, note that by Striling’s approximation for all \(m\in{\mathbb{N}}\) have \(\ln((2m)!)=(2m)\ln(2m)-m+O(\ln m)\). Therefore, for all \(n\) large enough \[\ln((2c_n)!)\geq 2c_n\ln(2c_n)-2c_n+O(\ln(c_n))\geq |t|nM_n/a_n+2c_n\ln (|t|nM_n/a_n)+|\ln\rho|\] where the inequality holds since \(n/a_n=O(\sqrt{n\ln(n)})\) and \(M_n=o(n^\varepsilon)\) for every \(\varepsilon>0\). This prove 8 .
Now, in order to complete the proof of the proposition it is enough to show that for all \(t\in{\mathbb{R}}\setminus\{0\}\), \[\label{Sh} \lim_{n\to\infty}\sum_{k=1}^{c_n-1}(-1)^k\sum_{A\subset N_n:\,|A|=2k}{\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell/a_n)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell/a_n)\right]=0.\tag{10}\] Let us take some \(A\subset N_n\) such that \(|A|=2k\) for \(1\leq k\leq c_n-1\). By expanding the functions \(\cos(x)\) and \(\sin(x)\) and denoting \(A^c=N_n\setminus A\) and \({\mathbb{N}}_0={\mathbb{N}}\cup\{0\}\), \[\label{R32n} R_{t,n}(A):={\mathbb{E}}\left[\prod_{\ell\in A}\sin(t{\theta}_\ell/a_n)\prod_{\ell\in N_n\setminus A}\cos(t{\theta}_\ell/a_n)\right]=\tag{11}\] \[\sum_{(m_\ell)\in{\mathbb{N}}^A}\,\sum_{(s_d)\in {\mathbb{N}}_0^{A^c}}(-1)^{\sum_{\ell\in A}(m_\ell+1)+\sum_{d\in A^c}s_d}\frac{t^{\sum_{\ell\in A}(2m_\ell-1)+2\sum_{d\in A^c}s_d}}{\prod_{\ell\in A}(2m_\ell-1)!\prod_{d\in A^c}(2s_d)!}{\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell^{2m_\ell-1}\prod_{d\in A^c}{\theta}_{d}^{2s_d}\right].\] Next, we note that since \(2s_d\) and \(2(m_\ell-1)\) are even and \(X_p\) are symmetric and independent, \[{\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell^{2m_\ell-1}\prod_{s\in A^c}{\theta}_{d}^{2s_d}\right]=0\] if and only if \[{\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell\right]=0.\] Indeed, both expectations vanish if and only if each prime \(p\) that divides some \(\ell\in A\) divides an even number of \(\ell\)’s in \(A\). Therefore, \[\left({\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell\right]=0\right)\Longrightarrow\left(R_{t,n}(A)=0\right).\]
Let us take some \(k\leq c_n-1\) and \(A\subset N_n=\{1,...,n\}\) such that \(|A|=2k\). We call the set \(A\) good if either \(A\) is not contained in \(\text{SF}\) (square free numbers) or there exists at least one prime number \(p\) that divides an odd amount of members \(\ell\) of \(A\). Then if \(A\) is good we have \[{\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell\right]=R_{t,n}(A)=0\] since \(\prod_{\ell\in A}{\theta}_\ell\) includes \(X_p\) with some odd power, unless \(A\) is not contained in \(\text{SF}\), but in this case \({\theta}_\ell=0\) for some \(\ell\) and so the above expectation anyway vanishes. Note that for \(k=1\) all subsets of \(N_n\) are good. Henceforth we assume that \(k>1\). Next, when \(k>1\) we call a set \(A\) bad if it is not good. Namely, if \(A\subset N_n\cap\text{SF}\) and every prime number \(p\) divides an even number of members \(\ell\in A\). Let us take a bad set \(A\subset N_n\cap\text{SF}\) with \(|A|=2k, \,1<k<c_n\). Next, we claim that there exists a constant \(d>0\) such that for all \(n\) large enough and \(1<k<c_n\) we have \[\label{A} \text{the number of bad sets among the sets of size }\, 2k\,\text{ does not exceed }\,d^k\left(\frac{n}{\ln(n)}\right)^{2k}.\tag{12}\] Let us first complete the proof of the proposition relying on this claim. By 6 and 9 , \[|R_{t,n}(A)|\leq (M_n|t|/a_n)^{2k}{\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell\in N_n\setminus A}{\theta}_\ell^2}].\] Notice that \[{\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell\in N_n\setminus A}{\theta}_\ell^2}]\leq {\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2}]e^{2k\frac{t^2}{4a_n^2}M_n}.\] Moreover, when \(k\leq c_n\) then \(2kM_n/a_n^2\leq 2c_nM_n/a_n^2\leq n^{1/2+\varepsilon}\ln(n)M_n/a_n^2\to 0\) as \(n\to\infty\) and so there is a constant \(C>0\) such that \[{\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell\in N_n\setminus A}{\theta}_\ell^2}]\leq C{\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2}]\] Therefore, using again that \(\binom{n}{2k}\leq \frac{n^{2k}}{(2k)!}\) and 12 we see that \[\left|\sum_{k=1}^{c_n-1}(-1)^k\sum_{A\subset N_n: \,|A|=2k}R_{t,n}(A)\right|=\left|\sum_{k=1}^{c_n-1}(-1)^k\sum_{A\subset N_n: \,|A|=2k,\,A\,\text{ is bad}}\,R_{t,n}(A)\right|\] \[\leq C{\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2}]\cdot\sum_{k=1}^{c_n-1}\frac{((\ln n)^2M_n|td|/a_n)^{2k}}{(2k!)}\leq {\mathbb{E}}[e^{-\frac{t^2}{4a_n^2}\sum_{\ell=1}^n{\theta}_\ell^2}]e^{(\ln n)^2M_n|td|/a_n}\to 0\text{ as }n\to\infty\] where we used that \((\ln n)^2M_n/a_n\to 0\) as \(n\to\infty\). Thus 10 follows, and the proof of the proposition is complete.
Proof of 12 Notice that every square free number \(\ell\leq n\) can be a product of at most \(\frac{\ln(n)}{\ln(2)}\) primes. Indeed, if \(\ell=p_1\cdots p_d\) then \(n\geq\ell\geq 2^d\). Therefore the number of primes that divide some \(\ell\in A\) does not exceed \(\frac{2k\ln(n)}{\ln 2}\) which does not exceed \(4c_n\ln(n)=O(n^{1/2+\varepsilon}(\ln (n))^{3/2}), \varepsilon<1/4\). On the other hand by the prime numbers theorem the number of primes inside \(N_n\) is of order \(\frac{n}{\ln(n)}\). Thus for all \(n\) large enough we can form at least \(\alpha_{k,n}:=(\frac{n}{2\ln(n)})^{2k}\) good subsets of \(N_n\) with cardinality \(2k\) simply by removing some member \(\ell\) of \(A\) and replacing it by one of the primes that does not divide all \(\ell\in A\). Let us denote the collection of good sets associated with a bad set \(A\) by \(G_A=\{G_{A,i}: 1\leq i\leq \alpha_{k,n}\}\). Let us denote the subcollection \(\tilde{G}_A\subset G_A\) that is formed from \(G_A\) by removing from \(G_A\) all sets in the collection \(G_B\) for all bad sets \(B\not=A\). We claim that the number of sets in the new collection \(\tilde{G}_A\) is at least \(q\alpha_{k,n}\) for some \(q>0\) (and all \(n\) large enough). Indeed, if for some \(B\not=A\) and some \(i,j\) we have \(G_{A,i}=G_{B,j}\) then \(A\) and \(B\) be can differ by at most two point. In the following argument is will increase clarity if we view \(A=(a_1,...,a_{2k})\) and \(B=(b_1,...,b_{2k})\) as vectors with different coordinates taking values in \(N_n\). By relabeling, for a fixed \(A\) we can always assume that \(a_{j}=b_j\) for all \(j>2\) and \(a_1\not=b_1\). Indeed, this has zero affect on the sets in the collections \(G_{A}\) and \(G_B\) (and we can do that simultaneously for all \(B\not=A\) such that \(G_A\cap G_B\not=\emptyset\)). Now, since \(a_1\not=b_1\) all the sets in \(G_B\) that are formed by replacing one of the coordinates \(b_j, j>1\) by some prime number as above cannot be one of the sets in \(G_A\) that are generated by replacing a coordinate \(a_m, m>1\) by some prime number. Therefore, the intersection of the union \(\cup_{B\not=A}G_B\) with \(G_A\) does not exceed \(\frac{n}{2\ln n}\), which is the number of sets that are formed by replacing only the first coordinate . We thus conclude that the collection \(\tilde{G}_A\) contains at least \(q\alpha_{k,n}\) sets for some constant \(q>0\) (and all \(n\) large enough and \(k\leq c_n\)). Since the collections \(\tilde{G}_A, A\text{ is bad}\) are disjoint we conclude that the number of the bad sets among the subsets of \(N_n\) with size \(2k\) is at most \(d^{k}\frac{\binom{n}{2k}\ln^{2k}(n)}{n^{2k}}\) for some constant \(d>0\), namely that 12 holds. ◻
For the sake of contradiction, let us assume that (possibly along a subsequence) \(S_n/a_n\) converges in distribution to some random variable \(Z\). Then for every real \(t\), \[\lim_{n\to\infty}{\mathbb{E}}[e^{itS_n/a_n}]={\mathbb{E}}[e^{itZ}].\] Note that due to the convergence of \(S_n/a_n\), Lemma 2 ensures that \(\liminf_{n\to\infty}\frac{a_n^2}{l_n}>0\). Thus by Proposition 3, for all \(t\in(-\delta_0,\delta_0)\setminus\{0\}\) we have \[\label{upp} \lim_{n\to\infty}{\mathbb{E}}[\cos(tS_n/a_n)]=0.\tag{13}\] Let us write \[{\mathbb{E}}[e^{itS_n/a_n}]={\mathbb{E}}[\cos(tS_n/a_n)]+i{\mathbb{E}}[\sin(tS_n/a_n)].\] Next, by 13 and since \(S_n/a_n\to Z\), for every real \(t\), \[{\mathbb{E}}[e^{itZ}]=i\lim_{n\to\infty}{\mathbb{E}}[\sin(tS_n/a_n)].\] Let us take some \(\varepsilon>0\). Then by the continuity of the characteristic function of \(Z\), for every sufficiently small nonzero \(t\) we have \[|{\mathbb{E}}[e^{itZ}]-1|<\varepsilon.\] Thus for all \(n\) large enough we have \[1+|{\mathbb{E}}[\sin(tS_n/a_n)]|^2=|i{\mathbb{E}}[\sin(tS_n/a_n)]-1|^2<(1+\varepsilon)^2\] which implies that \(|{\mathbb{E}}[\sin(tS_n/a_n)]|^2<2\varepsilon+\varepsilon^2\). Therefore, \[\lim_{n\to\infty}{\mathbb{E}}[\sin(tS_n/a_n)]=0\] and so by ?? we get that for every nonzero \(t\), \[{\mathbb{E}}[e^{itZ}]=\lim_{n\to\infty}{\mathbb{E}}[e^{itS_n/a_n}]=0.\] However for all \(t\) small enough \(|{\mathbb{E}}[e^{itZ}]|>\frac{1}{2}\) since characteristic functions are always continuous. 0◻
Taking \(a_n=1\) in 11 and using 5 we see that for all \(r\in{\mathbb{N}}\), \[{\mathbb{E}}[(S_n)^{2r}]=\sum_{k=1}^{[n/2]}(-1)^k\sum_{A\subset N_n:\,|A|=2k}\Gamma_{k,r}(A)\] where \[\Gamma_{k,r}(A)=\sum_{\substack{(s_d)\in{\mathbb{N}}_0^{A^c}, (m_\ell)\in{\mathbb{N}}^A\\ \sum_{\ell\in A}(2m_\ell-1)+2\sum_{d\in A^c}s_d=2r}} \frac{(2r)!}{\prod_{\ell\in A}(2m_\ell-1)!\prod_{d\in A^c}(2s_d)!}{\mathbb{E}}\left[\prod_{\ell\in A}{\theta}_\ell^{2m_\ell-1}\prod_{d\in A^c}{\theta}_{d}^{2s_d}\right].\] Notice that \(\Gamma_{k,r}(A)=0\) if \(k>r\) since \(k\leq \sum_{\ell\in A}(2m_\ell-1)\leq r\). Thus, for all \(n\) large enough \[\label{MomEq} {\mathbb{E}}[(S_n)^{2r}]=\sum_{k=1}^{r}(-1)^k\sum_{A\subset N_n:\,|A|=2k}\Gamma_{k,r}(A)\tag{14}\] Now, arguing like at the end of the proof of Proposition 3 for all \(n\) large enough we see that \(\Gamma_{k,r}(A)\not= 0\) only for \(O((\ln(n))^{2r} n^{r})\) sets \(A\). Therefore, \[{\mathbb{E}}[(S_n)^{2r}]\leq n^r(\ln(n))^{2r}\left(\max(1,M)\right)^{\max_{\ell\leq n}{\omega}(\ell)}\leq n^r(\ln(n))^{2r}\left(\max(1,M)\right)^{c\frac{\ln(n)}{\ln(\ln(n))}}.\] 0◻